Question

A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.845 m and 2.79 kg, respectively. When the propellor rotates at 511 rpm (revolutions per minute), what is its rotational kinetic energy?

Answer 1

m =2.79 kg, L = 0.845 m

$\omega$ =511 rev/min = 511x(2 $\pi$ /60)

$\omega$ = 53.48 rad/s

I = 5xML²/12

Rotatinal kinetic energy K.E =(1.2) $\omega$ ²

K.E =(1/2)(5xML²/12) $\omega$ ²

K.E =(1/2) [5x2.79x0.845x0.845/12](53.48)²

K.E = 1187 J

Answer 2

The propeller's rotational kinetic energy ?$K$ is

$$K=\frac{1}{2}I{\mathit{\omega }}^2$$

where ?$I$ is its moment of inertia with respect to the axis of rotation and ?$\mathit{\omega }$ is its angular speed.

The propeller consists of five identical uniform rods rotating about their ends. Its moment of inertia ?$I$ is therefore five times the moment of inertia of a single uniform rod of mass ?$m$ and length ?$L$ with respect to a perpendicular axis passing through its end.

$$I=5\left(\frac{1}{3}m{L}^2\right)=\frac{5}{3}m{L}^2$$

The rotational kinetic energy ?$K$ of the propeller is therefore

$$K=\frac{1}{2}\left(\frac{5}{3}m{L}^2\right){\mathit{\omega }}^2=\frac{5}{6}m{L}^2{\mathit{\omega }}^2$$

in terms of given quantities, although ?$\mathit{\omega }$ is given in rpm rather than in the appropriate SI unit of rad/s. Convert ?$\mathit{\omega }$.$$\mathit{\omega }=\left(515\text{ rev/min}\right)\left(\frac{2\mathit{\pi }\text{ rad}}{1\text{ rev}}\right)\left(\frac{1\text{ min}}{60\text{ s}}\right)=53.9\text{ rad/s}$$

Now, substitute the data in the expression for ?$K$ in order to obtain the numerical answer.

?=56(2.51 kg)(0.731 m)2(53.9 rad/s)2=3250 J